BigBurger

$\boxed{\textbf{True}},$ because rectangles have varying side lengths $a$ and $b$, and when $a=b$ the rectangle is a square. Therefore, rectangles are somtimes squares, but not always.

Given this recursion equation, we can use the given $a_0=1$ and $a_1 = 5$ conditions to solve for $a_2,$ and then $a_3$. If we first let $n=1,$ we get $$(a_1)^2 - a_0 a_2 = (-1)^1,$$

$$25 - 1(a_2) = -1,$$

$$a_2 = 26.$$

Using this same logic, try to figure out $a_3$ by substituting into the relation $n=2.$

The ratio of $\triangle ABC$ to $\triangle ACD$ to $\triangle ADE, $ which is just the ratio of the bases $BC$ to $CD$ to $DE$ (because the height for each of the triangles is the same). Therefore, we can write equations corresponding to these ratios. $$\frac{BC}{[ABC]} = \frac{DE}{[ADE]}, $$  $$\frac{5-CD}{24} = \frac{4-CD}{18}.$$

Solving this set of equations, we find $$90-18CD = 96-24CD,$$ $$6CD=6,$$ $$CD = 1.$$

Therefore, we can use this value of $CD$ to substitute back into the original equation. $$\frac{CD}{[ACD]} = \frac{DE}{[ADE]}, $$ $$\frac{1}{[ACD]} = \frac{3}{18}, $$ $$[ACD] = 6.$$

Thus, our answer is $\boxed{[ACD] = 6}.$

(Note: The diagram is not drawn to scale).

Since each person forwards to email to 3 other people, the expression for the number of additional people who received it after $n$ rounds (including Elijah) is $3^0 + 3^1 + 3^2 + 3^3 + \dots + 3^n. $ We can rewrite this as $\boxed{\sum_{k=0}^n 3^n}.$

The sum of the angles in a triangle is $180°, $ and we find that $$(180-106) + 45+ 31+ x = 180°.$$ We know this because vertical angles are equal, so the vertical angle to the angle of $x°$ is also equal to $x°.$ 

Now you can try to solve this equation yourself.

The area of the triangular pyramid is $\frac{1}{3} bh. $ We can make the base the top face, and the base has an area of $\frac{3 \cdot 4}{2} = 6.$ The height that is given is $\frac{1}{2} \cdot 4 = 2.$ Now, we can plug these values in, and we get $$V = \frac{1}{3} bh. $$ $$V = \frac{1}{3} (6)(2)$$ $$\boxed{V = 4 \text{ in}^3}.$$

The domain is the interval of the possible values of $x$. As we can see from the graph, the domain is infinity because the parabola extends infinitely. 

The range is the interval of the possible values of $y$. In this case, we see that the $y$ values cannot go below $-2, $ but they extend towards positive infinity. 

Therefore, the domain is $\boxed{(-\infty, +\infty)},$ and the range is $\boxed{(-2, +\infty)}. $

We see that we need to find $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}$, which simplifies to $\frac{a+b+c}{abc}.$ From here, we can now proceed with Vieta's. By Vieta's theorem (if we assume that the polynomial $f(x) = Ax^3 + Bx^2 + Cx + D$, we know that $$a+b+c = -\frac{B}{A}, $$ $$ab+ bc + ac = \frac{C}{A},$$ $$abc = -\frac{D}{A},$$ where $A=1, B=-6, C=13, D=-18.$ Therefore, we get that $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac} = \frac{-6}{-18} = \boxed{\frac{1}{3}}.$

If we add the equations given in the question, we get $4x+4y+4z = 1400. $ Dividing by $4$ on each side, we get $x+y+z=350.$ Therefore, the average of $x,y,z$ is just $\frac{x+y+z}{3} = \boxed{\frac{350}{3}}.$

Since $a,b,c$ all have to be natural numbers, $a=1$ because it must be less than $\frac{10}{7}$, but also an integer greater than $0$. Furthermore, we now have $$\frac{1}{b+\frac{1}{c}} = \frac{3}{7}$$ $$\frac{c}{bc+1} = \frac{3}{7}$$. Therefore, we find that $c=3$ and $b=2$. Thus, our answer is $(a,b,c) = \boxed{(1,2,3)}.$

We first focus on the second part of the statement. We find that $(1/2 \text{cm})^2 = \frac{1}{4} \text{cm}^2. $ Therefore, we get that $\frac{\frac{1}{4} \text{cm}^2}{\frac{1}{2} \text{cm}^2} = \boxed{\frac{1}{2}}.$

3. (a) We know that $a_3 = 45$ and $a_5 = 101.25.$ If we express $a_3 = a_1 r^2$ and $a_5 = a_1 r^4$, we can find the common ratio if we divide $a_5/a_3$. Doing this, we get $$r^2 = \frac{a_5}{a_3},$$ $$r^2 = \frac{101.25}{45} = 2.25. $$ Since we know that all of the terms are positive, the common ratio $r=1.5,$ and we find the first term by substituting the common ratio into the equation for $a_3$. We get $$45 = a_1 (1.5)^2,$$ $$a_1 =20. $$ Therefore, our answer is $\boxed{a_1 = 20, r=1.5}.$

    (b) We find the largest term greater than $2000$ by expressing $a_x = a_1 r^{x-1}. $ Given that $r=1.5$ (from part(a)), we find that $x-1 = 12$. Therefore, the thirteenth term exceeds $2000$, and $\boxed{a_{13} \approx 2594.93}. $

5. (a) We know three successive terms of a geometric sequence are $x-6,x,2x+9$. We can express these terms like we did before, and we get $$x-6 = a r,$$ $$x = ar^2,$$ $$2x+9 = ar^3.$$ Therefore, we can solve two equations for $r$, and we get $$r = \frac{x}{x-6} = \frac{2x+9}{x}.$$ Solving for $x$, we get $(x-9)(x+6)=0$. Since all terms in the sequence must be positive, our answer is $\boxed{x=9}$. 

    (b) Since $x$ is the $5$th term, $a_5 = x = a_1 r^4.$ We can solve for $r$ from the equations before, where $r = \frac{x}{x-6}. $ We know that $x=9$, so $r=\frac{9}{3} = 3.$ Therefore, we know $9 = a_1 (3)^4. $ Therefore, we get $\boxed{a_1 = \frac{1}{9}}.$

6. (a) Since Nurhaniza was her salary increase $12$ percent every year, her monthly salary after 6 years would just be $$\text{final salary = starting salary }\cdot \text{(annual increase %)}^6.$$ Plugging in our values for the starting salary and the annual increase %, we get $$\text{final salary} = RM1500 \cdot (112 \text{%})^6, $$ $$\text{final salary} = RM2960.73. $$

    (b) Part (b) asks us for the number of years that she must stay at the company until her monthly salary exceeds $RM3500.$ Therefore, we are asked to find $n$ in the inequality $ RM1500 \cdot (112 \text{%})^n > RM3500.$ Solving this inequality, we find that $n=8.$ Therefore, it takes $\boxed{8}$ years with the company for Nurhaniza's monthly salary to exceed $RM3500.$

Hope this was helpful!

We can solve this problem using Vieta's theorem. We want to find $\frac{1}{a} + \frac{1}{b}$, which simplifies to $\frac{b+a}{ab}$. Using Vieta's, we know that $a+b = -\frac{b}{a}$ and $ab = \frac{c}{a}$, where $a,b,c$ refer to a quadratic $ax^2+bx+c.$ Therefore, $a=5, b=-11, c=4. $ Using this, we find that $a+b = \frac{11}{5}$, and $ab = \frac{4}{5}$. Thus, we can find that $\frac{1}{a} + \frac{1}{b} = \boxed{\frac{11}{4}}.$

We know that $\triangle JKL \sim \triangle JMN$ by $AA$ similarity ($\angle JKL \cong \angle JMN$ and $\angle JLK \cong \angle JNM$ because $KL \parallel MN$). Therefore,

$$\frac{JK}{JM} = \frac{JL}{JN},$$

$$\frac{36}{JM} = \frac{48}{60}, $$

$$JM = 45. $$

Given that $JM=45$, we see that $MK = JM - JK$, and $JK=36$ using the similarity ratio above. Thus, $MK = 45 - 36 = \boxed{9}.$

First, we can graph the parabola by graphing $y=(x-2)^2+1.$ The vertex of the parabola is $(2,1)$ and the $y$-intercept is where $x=0$. This turns out to be $y=(-2)^2+1 = 5$, so the $y$-intercept is $(0,5)$. Therefore, we can use this information to graph the parabola. We find the solution set of this by setting a test point, $(0,0).$ If this test point satisfies the inequality, then the region in which the test point is included is the solution set. I'll let you take it from here. 

If we have a polynomial with variable $x$, we can write the polynomial in factored form given its roots. Therefore, we get 

$$(x-(-2))(x-3)(x-4).$$ When we expand this out, we get $$x^3-5x^2-2x+24.$$ When we look at the other condition in the problem, we must have an $y$-intercept with a value of $24$. The $y$-intercept is where $x=0$, so we just look at the constant term in our polynomial and see that it is $24$. Thus, the least degree polynomial is $\boxed{x^3-5x^2-2x+24}.$